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Sulphur Trioxide

Chlorine is p...

Question

Chlorine is prepared in the laboratory by treating manganese dioxide $(M n O_{2})$ with aqueous hydrochloric acid according to the reaction

$4 H C l (a q) + M n O_{2} (s) \to 2 H_{2} O (l) M n C l_{2} (a q) + C l_{2} (g)$

How many grams of $H C I$ reacts with $5.0 g$ of manganese dioxide?

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Solution

Amount of manganese oxide $= 5 g$

We have to find amount of $H C l$ required.
Molar mass of $M n O_{2} : 55 + 32 = 87 g m o l^{- 1}$
Molar mass $H C l : 1 + 35.5 = 36.5 g$

According to the balanced chemical equation,
$1 mol of M n O_{2} reacts with 4 mol of HCI$

So, $87 g of M n O_{2} reacts with (36.5 \times 4) g = 146 g of HCl.$
So, $5 g of M n O_{2} will react with$

$= \frac{146 g}{87 g} \times 5 g of HCl$
$= 8.4 g HCl$

Therefore, $8.4 g of HCl will react with 5 g of M n O_{2} .$

Final answer: $8.4 g o f H C l w i l l r e a c t w i t h 5 g of M n O_{2}$

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chlorine is prepared in the lab by treating manganese dioxide (MNO2) wiht aqueous hychloric acid according to the equation

4HCl(aq) + MnO2 (s)------------- 2H20 (l) + MnCl2 (aq) + Cl2(g)

how many grams of HCl react with 5g of manganese dioxide?

Chlorine is prepared in the lab by treating with MnO2 with aqueous HCl

How many grams of HCl reacts with 5g of MnO2?

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